How maby grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at `268K`?
`(K_(f)` for water is `1.86 K kg mol^(-1))`

Options

A. `899.04g`
B. `886.02g`
C. `868.06g`
D. `880.07g`

Explanation

To solve the problem of how many grams of methyl alcohol should be added to a 10-liter tank of water to prevent it from freezing at 268 K, we will follow these steps:

### Step 1: Understand the given data
- Freezing point of pure water (T0) = 273 K
- Freezing point of the solution (Tf) = 268 K
- Depression in freezing point (ΔTf) = T0 - Tf = 273 K - 268 K = 5 K
- Cryoscopic constant (Kf) for water = 1.86 K kg mol^(-1)
- Volume of water = 10 liters

### Step 2: Convert the volume of water to mass
Since the density of water is approximately 1 kg/L, the mass of water (solvent) can be calculated as:
\[ \text{Mass of water} = 10 \, \text{L} \times 1 \, \text{kg/L} = 10 \, \text{kg} \]

### Step 3: Use the formula for depression in freezing point
The relationship between the depression in freezing point (ΔTf), the cryoscopic constant (Kf), and the molality (m) of the solution is given by:
\[ \Delta Tf = Kf \times m \]
Rearranging this gives:
\[ m = \frac{\Delta Tf}{Kf} \]

### Step 4: Calculate the molality
Substituting the known values:
\[ m = \frac{5 \, \text{K}}{1.86 \, \text{K kg mol}^{-1}} \]
\[ m \approx 2.688 \, \text{mol/kg} \]

### Step 5: Calculate the number of moles of methyl alcohol required
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Therefore, the number of moles (n) of methyl alcohol needed can be calculated as:
\[ n = m \times \text{mass of solvent (kg)} \]
\[ n = 2.688 \, \text{mol/kg} \times 10 \, \text{kg} \]
\[ n \approx 26.88 \, \text{mol} \]

### Step 6: Calculate the mass of methyl alcohol
The molar mass of methyl alcohol (methanol, CH3OH) is approximately 32 g/mol. Thus, the mass (W) of methyl alcohol required can be calculated as:
\[ W = n \times \text{molar mass} \]
\[ W = 26.88 \, \text{mol} \times 32 \, \text{g/mol} \]
\[ W \approx 860.16 \, \text{g} \]

### Final Answer
To prevent the freezing of water at 268 K, approximately **860.16 grams** of methyl alcohol should be added to the 10-liter tank of water.

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